Problem 1 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Compute \(y^{\prime}(x)\) for a. \(\quad y=2 x^{3}-5\) b. \(y=\frac{2}{x^{2}}\) c. \(y=\frac{5}{(x+1)^{2}}\) d. \(y=\left(1+x^{2}\right)^{0.5}\) e. \(\quad y=\sqrt{1-x^{2}}\) f. \(y=\left(1-x^{2}\right)^{-0.5}\) g. \(\quad y=(2-x)^{4}\) h. \(y=\left(3-x^{2}\right)^{4}\) i. \(y=\frac{1}{\sqrt{7-x^{2}}}\) j. \(y=\left(1+(x-2)^{2}\right)^{2}\) k. \(y=(1+3 x)^{1.5} \quad\) l. \(\quad y=\frac{1}{\sqrt{16-x^{2}}}\) \(\mathrm{m} . \quad y=\sqrt{9-(x-4)^{2}}\) n. \(y=\left(9-x^{2}\right)^{1.5}\) o. \(y=\sqrt[3]{1-3 x}\)

Step-by-Step Solution

Answer

The derivatives are: a) 6x², b) -4/x³, c) -10/(x+1)³, d) x/√(1+x²), e) -x/√(1-x²), f) x/(1-x²)^{1.5}, g) -4(2-x)³, h) -8x(3-x²)³, i) x/(7-x²)^{1.5}, j) 4(x-2)(1+(x-2)²), k) 4.5(1+3x)^{0.5}, l) x/(16-x²)^{1.5}, m) (x-4)/√(9-(x-4)²), n) -3x(9-x²)^{0.5}, o) -1/(1-3x)^{2/3}.

Step 1: Differentiate part a

The function given is \( y = 2x^{3} - 5 \). To find the derivative \( y'(x) \), apply the power rule. The derivative of \( 2x^3 \) is \( 6x^2 \) and the derivative of constant \( -5 \) is \( 0 \), so \( y'(x) = 6x^2 \).

Step 2: Differentiate part b

For \( y = \frac{2}{x^2} = 2x^{-2} \), use the power rule. The derivative is \( y'(x) = -4x^{-3} = -\frac{4}{x^3} \).

Step 3: Differentiate part c

For \( y = \frac{5}{(x+1)^2} = 5(x+1)^{-2} \), use the power rule. The derivative is \( y'(x) = -10(x+1)^{-3} = -\frac{10}{(x+1)^3} \).

Step 4: Differentiate part d

Given \( y = (1+x^2)^{0.5} \), use the chain rule: \( u = 1+x^2 \), \( u' = 2x \), so \( y'(x) = 0.5(1+x^2)^{-0.5} \, * \, 2x = \frac{x}{\sqrt{1+x^2}} \).

Step 5: Differentiate part e

For \( y = \sqrt{1-x^2} = (1-x^2)^{0.5} \), use the chain rule: \( u = 1-x^2 \), \( u' = -2x \), hence \( y'(x) = 0.5(1-x^2)^{-0.5} \, * \, (-2x) = \frac{-x}{\sqrt{1-x^2}} \).

Step 6: Differentiate part f

Given \( y = (1-x^2)^{-0.5} \), apply the chain rule: \( u = 1-x^2 \), \( u' = -2x \), thus \( y'(x) = -0.5(1-x^2)^{-1.5} \, * \, (-2x) = \frac{x}{(1-x^2)^{1.5}} \).

Step 7: Differentiate part g

The function is \( y = (2-x)^4 \). Using the chain rule where \( u = 2-x \) and \( u' = -1 \), we have \( y'(x) = 4(2-x)^3 \, * \, (-1) = -4(2-x)^3 \).

Step 8: Differentiate part h

For \( y = (3-x^2)^4 \), use the chain rule: \( u = 3-x^2 \), \( u' = -2x \), so \( y'(x) = 4(3-x^2)^3 \, * \, (-2x) = -8x(3-x^2)^3 \).

Step 9: Differentiate part i

The function is \( y = \frac{1}{\sqrt{7-x^2}} = (7-x^2)^{-0.5} \). Apply chain rule: \( u' = -2x \), so \( y'(x) = -0.5(7-x^2)^{-1.5} \, * \, (-2x) = \frac{x}{(7-x^2)^{1.5}} \).

Step 10: Differentiate part j

Given \( y = (1+(x-2)^2)^2 \), employ the chain rule: \( u = 1+(x-2)^2 \), \( u' = 2(x-2) \), hence \( y'(x) = 2(1+(x-2)^2) \, * \, 2(x-2) = 4(x-2)(1+(x-2)^2) \).

Step 11: Differentiate part k

For \( y = (1+3x)^{1.5} \), apply the chain rule: \( u = 1+3x \), \( u' = 3 \), thus \( y'(x) = 1.5(1+3x)^{0.5} \, * \, 3 = 4.5(1+3x)^{0.5} \).

Step 12: Differentiate part l

The function is \( y = \frac{1}{\sqrt{16-x^2}} = (16-x^2)^{-0.5} \). Using the chain rule: \( u' = -2x \), so \( y'(x) = -0.5(16-x^2)^{-1.5} \, * \, (-2x) = \frac{x}{(16-x^2)^{1.5}} \).

Step 13: Differentiate part m

For \( y = \sqrt{9-(x-4)^2} = (9-(x-4)^2)^{0.5} \), use chain rule: \( u = 9-(x-4)^2 \), \( u' = -2(x-4) \), so \( y'(x) = 0.5(9-(x-4)^2)^{-0.5} \, * \, (-2(x-4)) = \frac{(x-4)}{\sqrt{9-(x-4)^2}} \).

Step 14: Differentiate part n

Given \( y = (9-x^2)^{1.5} \), apply chain rule: \( u = 9-x^2 \), \( u' = -2x \), so \( y'(x) = 1.5(9-x^2)^{0.5} \, * \, (-2x) = -3x(9-x^2)^{0.5} \).

Step 15: Differentiate part o

The function is \( y = \sqrt[3]{1-3x} = (1-3x)^{1/3} \). Using the chain rule: \( u = 1-3x \), \( u' = -3 \), so \( y'(x) = \frac{1}{3}(1-3x)^{-2/3} \, * \, (-3) = -\frac{1}{(1-3x)^{2/3}} \).

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus, used when you need to differentiate composite functions. A composite function is a combination of two or more functions, where one function is nested inside another. To apply the chain rule, you differentiate the outer function, keeping the inner function intact, and then multiply it by the derivative of the inner function.

The formula for the chain rule is:\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]Here, \(y\) is a function of \(u\), and \(u\) is a function of \(x\). This approach lets you tackle complex differentiation problems step-by-step, making it easier to manage.

For example, consider the function \(y = (1 + x^2)^{0.5}\). To differentiate using the chain rule:

• Let \(u = 1 + x^2\), then \(y = u^{0.5}\).
• The derivative of \(y\) with respect to \(u\) is \(\frac{dy}{du} = 0.5u^{-0.5}\).
• The derivative of \(u\) with respect to \(x\) is \(\frac{du}{dx} = 2x\).
• Apply the chain rule: \(\frac{dy}{dx} = 0.5(1+x^2)^{-0.5} \cdot 2x = \frac{x}{\sqrt{1+x^2}}\).

This methodology is crucial when simplifying and differentiating functions, allowing us to dive deeper into the behavior of functions and their graphs.

Power Rule

The power rule is a practical tool in differentiation, making it simple to find the derivatives of functions where the variable is raised to a power. The rule states that if you have a function in the form \(y = x^n\), where \(n\) is a constant, then the derivative \(y'\) is given by \(nx^{n-1}\).

For a better understanding, consider this example: if \(y = 2x^3\), applying the power rule gives \(y' = 3 \cdot 2x^{3-1} = 6x^2\). It's straightforward and efficient, making it suitable for any polynomial expression.

• The power rule also extends to negative and fractional exponents.
• For instance, if \(y = x^{-2}\), then \(y' = -2x^{-3}\).
• In fractional form, such as \(y = x^{1/2}\), the derivative becomes \(y' = \frac{1}{2}x^{-1/2}\).

This rule helps streamline the process of differentiation, allowing for quick analysis and comprehension of polynomial trends and behaviors in function graphs.

Functions and Graphs

Understanding functions and their graphs is crucial in calculus, as it helps visualize the impact of differentiation. Functions can be simple linear equations, polynomial expressions, or more complex shapes like trigonometric or exponential forms. Graphically, differentiation provides insight into the slope of the tangent lines at various points on a function's curve.

• The derivative tells us the rate of change of the function at any given point.
• A positive derivative indicates that the function is increasing, while a negative derivative shows it is decreasing.
• A zero derivative points to a potential maximum or minimum value, also known as critical points.

By differentiating simple functions like polynomials or fractions, we gain valuable information about the graph's behavior such as concavity and points of inflection.

For example, in the function \(y = (2-x)^4\), differentiation helps us understand the rate and direction of change as \(x\) varies. Through these insights, one can predict how the graph will look and identify any significant characteristics like symmetry or intercepts.

Analyzing functions through their derivatives is a powerful method to explore their graphical representation and underlying mechanics.